Let $I=[0,1]$ and $T:L^{1}(I)\rightarrow L^{1}(I)$ defined by $$T(f)(x)=\int_{0}^{x}f(t)dt, \quad f\in L^{1}(I),\: x\in I.$$
Show that $T$ is inyective.
Remark: If $f$ were a continuous function, then by the fundamental theorem of calculus we have that if $T(f)=0$, then $f=0$.
The problem is when $f$ is not continuous, in this case, I think that we can not show $f=0$, the most we can pretend is $f=0$ almost surely, but I have not been able to show it.
Added:
I want to share the following reasoning and know if it is correct, I appreciate the answers you have given so far.
Let $f\in L^{1}(I)$ suc that $Tf=0$, then for al $x\in I$ we have $\int_{0}^{x}f(t)dt=0$, then
Fact 1: For all interval $J\subset I$ we have $\int_{J}f(t)dt=0$.
Proof: Let $a,b \in I$ the extreme points of $J$ with $a<b$, then $$\int_{J}f(t)dt=\int_{a}^{b}f(t)dt=\int_{0}^{b}f(t)dt-\int_{0}^{a}f(t)dt=Tf(b)-Tf(a)=0.$$
But we know that
Theorem 1: Let $f:[a,b]\rightarrow \mathbb{R}$ be a bounded function, then $f$ is Riemann integrable if and only if $\lambda\left(\left\{x\in[a,b]\: : \: f \mbox{ is discontinuous in }x\right\}\right)=0$ where $\lambda$ is Lebesgue measure in $I$.
Since $f\in L^{1}(I)$ then $f$ is Riemann integrable in $I$, furthermore, $f$ is Riemann integrable in $J$ for all $J\subseteq I$. Also by the definition of integral of Riemann we must have $f$ bounded in $I$. Therefore, $f$ is continuous $\lambda$ almost everywhere in $I$.
Let $x\in I$ such that $f$ is contiuous in $x$, then there exists an inteval $J\subset I$ with $x\in J$ such that $f $ is continuous in $J$, we consider $a,b\in I$ the extreme points of $J$ with $a<b$, then we consider
$F:[a,b]\rightarrow \mathbb{R}$ defined by $F(y)=\int_{a}^{y}f(t)dt$ for all $y\in J$, since $f$ is continuous in $J$, then $F'(y)=f(y)$.
Thus, for each $y\in J$ we conisder the interval $J_{y}\subset J$ with extreme points $a$ and $y$, then by Fact 1 we have $$0=\int_{J_{y}}f(t)dt=\int_{a}^{y}f(t)dt=F(y).$$ Then $F(y)=0$ for all $y\in J$, then $f(y)=F'(y)=0$ for all $y\in J$, in particular, $f(x)=0$.
Therefore, $f=0$ $\lambda$ almost everywhere in $I$.
Hint: Suppose $T(f)=0$. Let $A=\{x\mid f(x)>\frac{1}{n}\}$. Show that $\int_A f\,\textrm{d}\mu=0$, so $\mu(A)=0$.